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3 years ago

2 characteristics and 2 differences for spiral and elliptical galaxies

2 answers:

5 0

Spiral galaxies have three main components: a bulge, disk, and halo (see right). The bulge is a spherical structure found in the center of the galaxy. This feature mostly contains older stars. The disk is made up of dust, gas, and younger stars. The disk forms arm structures. Our Sun is located in an arm of our galaxy, the Milky Way. The halo of a galaxy is a loose, spherical structure located around the bulge and some of the disk. The halo contains old clusters of stars, known as globular clusters.

Elliptical galaxies are shaped like a spheriod, or elongated sphere. In the sky, where we can only see two of their three dimensions, these galaxies look like elliptical, or oval, shaped disks. The light is smooth, with the surface brightness decreasing as you go farther out from the center. Elliptical galaxies are given a classification that corresponds to their elongation from a perfect circle, otherwise known as their ellipticity. The larger the number, the more elliptical the galaxy is. So, for example a galaxy of classification of E0 appears to be perfectly circular, while a classification of E7 is very flattened. The elliptical scale varies from E0 to E7. Elliptical galaxies have no particular axis of rotation.

saul85 [17]3 years ago

4 0

Spiral galaxies are in a spiral like motion while eliptical galaxies are shaped like an egg. Spiral galaxies look like they have a core while eliptical galaxies dont.

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Alkenes: draw the product of 1-chloro-2-ethylcyclohexene with hydrogen gas and a platinum catalyst

atroni [7]

Reacting 1-chloro-2-ethylcyclohexene with hydrogen gas using a platinum catalyst would give a product of 1-chloro-2-ethylcyclohexane.

Hydrogen gas is a reducing agent, which in this reaction, simply mean that the alkene double bond in the cyclohexene will disappear because one of the two bonds forming the double bond (in the alkene) will be connected to a hydrogen atom. The platinum catalyst is necessary to allow the reaction to proceed at a much lower (activation) energy than would have been required.

7 0

3 years ago

Cell membranes are selectively permeable. This means that A. only water can move freely across the cell membrane. B. any substan

MaRussiya [10]

Answer:

C. some substances can move freely across the cell membrane, while others must be transported.

Explanation:

6 0

3 years ago

Hydrochloric acid reacts faster with powdered zinc than with an equal mass of zinc strips because the greater the surface area o

inysia [295]

Answer:

increases the frequency of particle collisions

Explanation:

One factor upon which the rate of reaction depends is the surface area of reactants.

According to the collision theory, reactions occur when reactant particles having the required (activation) energy collide with each other, this collision is inelastic. However, collision of particles having energies less than the activation energy results in elastic collisions and no chemical reaction.

The more the exposed surface area of reactants, the greater the number of particles that come into contact with each other and the more the chances of frequent effective collisions that lead to reaction.

Thus, powdered zinc reacts faster with hydrochloric acid than zinc strips

7 0

3 years ago

Suppose you want to prepare a buffer with a pH of 4.59 using formic acid. What ratio of [sodium formate]/[formic acid) do you ne

katrin [286]

Answer:

7.08

Explanation:

To solve this problem we'll use the Henderson-Hasselbach equation:

pH = pka + log $\frac{[A^-]}{[HA]}$

Where $\frac{[A^-]}{[HA]}$ is the ratio of [sodium formate]/[formic acid] and pka is equal to -log(Ka), meaning that:

pka = -log (1.8x10⁻⁴) = 3.74

We input the data:

4.59 = 3.74 + log $\frac{[A^-]}{[HA]}$

And solve for $\frac{[A^-]}{[HA]}$ :

0.85 = log $\frac{[A^-]}{[HA]}$

$10^{(0.85)}$ = $\frac{[A^-]}{[HA]}$

$\frac{[A^-]}{[HA]}$ = 7.08

3 0

3 years ago

Compared to the ideal bond angle, the actual/real bond angle in a molecule of water will be:______.

inna [77]

Answer: d. smaller.

Explanation:

Hybridization :

$N=\frac{1}{2}[V+N+A-C]$

N = number of electrons

where, V = number of valence electrons present in central atom i.e. oxygen = 6

N = number of monovalent atoms bonded to central atom=32

C = charge of cation = 0

A = charge of anion = 0

The number of electrons is 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral. The bond angle for tetrahedral geometry is $109.5^0$

But as there are two atoms around the central oxygen, the third and fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be bent and the bond angle will be smaller than expected and is $104.5^0$

8 0

3 years ago